Moving letters on the grid

/* Move the letter 2x3 = 6 squares with five letters ABCDE left empty in the lower right corner. 1.
 

The letters in the lattice adjacent to the blank lattice can be moved to the space. For example, C and E in the figure can be moved. The following figure shows the moving situation: a B d e ca B CD E for convenience, we use a string to represent the letters in the six grids. For example, the two situations above are represented: the requirements for AB * decabcd * E are as follows: Program The user inputs several strings that indicate the situation. The program outputs whether the initial status can be reached by moving several times. You can achieve output 1; otherwise, output 0. The initial status is ABCDE *. The format entered by the user is: an integer N, indicating that there are n rows following the status. The program output should also be n rows 1 or 0. For example, if the user input is 3abcde * AB * deccaed * B, the program should output: 110 */import Java. util. imports; import Java. util. arraylist; import Java. util. queue; import Java. util. using list; import Java. util. list; public class letter movement {// exchange element public static string swap (string S, int K, Int J) {char [] C = S. tochararray (); char T = C [k]; C [k] = C [J]; C [J] = T; return new string (c );} public static int F (string s) {list <string> Lis = ne W arraylist <string> (); // record all exchange results (check for repeated use) queue <string> queue = new queue list <string> (); // Queue (used to test all possibilities) int [] d = {-1,-3, 1, 3}; // defines the direction (left, top, right, bottom) queue. offer (s); // enter Lis. add (s); // Add record while (queue. size ()> 0) {string firststr = queue. poll (); // If (firststr. equals ("ABCDE *") return 1; // locate the result and exit int K = firststr. indexof ("*"); // locate the "*" position for (INT I = 0; I <4; I ++) {// left, top, right, int J = K + d [I] in the next four directions; If (j> = 0 & J <= 5) {// The current direction does not cross the border string T = swap (firststr, K, J); // exchange element if (! Lis. contains (t) {// Lis does not contain t queue. offer (t); // enter Lis. add (t); // Add record }}} return 0;} public static void main (string [] ARGs) {empty scan = new empty (system. in); system. out. println ("input integer N"); int n = scan. nextint (); scan. nextline (); // receives the carriage return string [] S = new string [N]; for (INT I = 0; I <n; I ++) {s [I] = scan. nextline (); // initial data} For (INT I = 0; I <n; I ++) {system. out. println (f (s [I]); // expected result }}}